My understanding is that since Current = Charges/Time. If there exists a resistance to the flow of charges, then that must mean the charges slow down, meaning that more time is required to pass through a point. So, the current should then decrease. But, since this opposition to the flow of charges doesn't exist in the ENTIRE circuit, it should really only decrease the current in the resistor, right?
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1$\begingroup$ You posed th same question already, why a second time? $\endgroup$– trulaSep 2, 2020 at 20:22
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$\begingroup$ Ah, that question is different. It's only that I was asking about the current obtained from the V=IR formula but a fellow here said that I should post a separate question about the current being lesser in a resistor. Didn't mean to spam! $\endgroup$– ihateelectricalphysicsSep 2, 2020 at 20:24
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5$\begingroup$ Does this answer your question? Is the current obtained using Ohm's law that which passes through the resistor or that which passes through the circuit? $\endgroup$– user12986714Sep 2, 2020 at 20:27
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$\begingroup$ No suitable explanation has been provided so far, unfortunately. Everyone seems to be talking about how the current through the resistor may be lesser or the same as compared to the current flowing through the circuit. No one seems to tell me if the current obtained via V=IR is that which flows in the resistor or in the circuit. Or perhaps, I may not be understanding it well yet. $\endgroup$– ihateelectricalphysicsSep 2, 2020 at 20:29
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$\begingroup$ Can you add a schematic of an example circuit of the type you want to ask about so we can discuss it more concretely? (Also, are you familiar with Kirchhoff's Current Law?) $\endgroup$– The PhotonSep 2, 2020 at 20:41
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2 Answers
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In circuit theory we have Kirchhoff's current law (KCL), which to a physicist is just a statement of conservation of charge:
The algebraic sum of currents in a network of conductors meeting at a point is zero.
(This is different from what you said in comments, "the sum of all the current in the entire circuit must be equal to zero." It applies at any single node in a circuit, not to the circuit as a whole)
This means if we consider a simple circuit with two resistors in series, like this,
then the sum of the currents flowing in to point B must be 0. Or, said another way, the current flowing in to point B must be equal to the current flowing out of point B.
Practically, it means the current flowing through resistor R1 to point B must be equal to the current flowing out of point B through R2.
A similar argument applies at point A. Any current flowing in to R1 from point A must be coming from current flowing out of the source V1.
Ultimately, in a circuit like this with only a single loop, the current in all elements must be equal.
answered Sep 3, 2020 at 4:04
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$\begingroup$ I think I've started to understand it a bit better: A fellow told me: Current has to be the same everywhere in a series circuit. This includes inside the resistor (which is a part of the circuit). If the current were not the same everywhere, then charge would be piling up somewhere. But a pile up of charge would lead to an electric field that would tend to get the charge moving again, smoothing out the current. So current is not less inside a resistor than outside. Instead, current decreases uniformly everywhere in the circuit when a resistor is part of it. $\endgroup$ Sep 3, 2020 at 4:06
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$\begingroup$ or perhaps: In the steady state, the current everywhere would settle down to a rate that is set the by time to get through the resistor, so that it’s no longer a bottleneck. No faster, and no slower. I think that’s the way to think about it. $\endgroup$ Sep 3, 2020 at 4:19
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The answers in your other question should tell you the answer to this one. the sum of the resistors in a circuit determine the magnitude of the current, which flows thru all parts of the circuit, The resistor limits the number of charges per second, so yes, if the resistance is greater, the charges are slower, but in all of the circuit, in the resistor they are faster than in the rest.
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$\begingroup$ the churches was some autocorrect, sorry it is edited now, thanks $\endgroup$– trulaSep 3, 2020 at 17:15